Difference between revisions of "1953 AHSME Problems/Problem 7"

(Solution)
(Previous page was a mess)
 
Line 11: Line 11:
 
== Solution ==
 
== Solution ==
  
Multiplying the numerator and denominator by <math>\sqrt{a^2+x^2}</math> results in <cmath>\frac{a^2+x^2-x^2+a^2}{(a^2+x^2)(\sqrt{a^2+x^2)}}=\frac{2a^2}{(a^2+x^2)(\sqrt{a^2+x^2)}}.</cmath> Since <math>\sqrt{a^2+x^2}=(a^2+x^2)^{\frac{1}{2}}</math>, if we call <math>a^2+x^2=p</math>, the denominator is really just <math>p^1\cdot{p^{\frac{1}{2}}}=p^{\frac{3}{2}}=(a^2+x^2)^ fraction is just </math>\boxed{\textbf{(D) } \frac{2a^2}{(a^2+x^2)^{\frac{3}{2}}}}$.
+
Multiplying the numerator and denominator by <math>\sqrt{a^2+x^2}</math> results in  
 +
<cmath>\frac{a^2+x^2-x^2+a^2}{(a^2+x^2)(\sqrt{a^2+x^2)}}=\frac{2a^2}{(a^2+x^2)(\sqrt{a^2+x^2)}}.</cmath>  
 +
Since <math>\sqrt{a^2+x^2}=(a^2+x^2)^{\frac{1}{2}}</math>,  
 +
the denominator is <math>(a^2+x^2)^2\cdot (a^2+x^2)^{\frac{1}{2}} = (a^2+x^2)^{\frac{3}{2}}</math> <math>\boxed{\textbf{(D) } \frac{2a^2}{(a^2+x^2)^{\frac{3}{2}}}}</math>.
 +
 
 +
==See Also==
 +
 
 +
{{AHSME 50p box|year=1953|num-b=6|num-a=8}}
 +
 
 +
[[Category:Introductory Algebra Problems]]
 +
{{MAA Notice}}

Latest revision as of 20:40, 1 April 2017

Problem

The fraction $\frac{\sqrt{a^2+x^2}-\frac{x^2-a^2}{\sqrt{a^2+x^2}}}{a^2+x^2}$ reduces to:

$\textbf{(A)}\ 0 \qquad \textbf{(B)}\ \frac{2a^2}{a^2+x^2} \qquad \textbf{(C)}\ \frac{2x^2}{(a^2+x^2)^{\frac{3}{2}}}\qquad \textbf{(D)}\ \frac{2a^2}{(a^2+x^2)^{\frac{3}{2}}}\qquad \textbf{(E)}\ \frac{2x^2}{a^2+x^2}$

Solution

Multiplying the numerator and denominator by $\sqrt{a^2+x^2}$ results in \[\frac{a^2+x^2-x^2+a^2}{(a^2+x^2)(\sqrt{a^2+x^2)}}=\frac{2a^2}{(a^2+x^2)(\sqrt{a^2+x^2)}}.\] Since $\sqrt{a^2+x^2}=(a^2+x^2)^{\frac{1}{2}}$, the denominator is $(a^2+x^2)^2\cdot (a^2+x^2)^{\frac{1}{2}} = (a^2+x^2)^{\frac{3}{2}}$ $\boxed{\textbf{(D) } \frac{2a^2}{(a^2+x^2)^{\frac{3}{2}}}}$.

See Also

1953 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 6
Followed by
Problem 8
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50
All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS